(a) Interpretation: The mass of the dopant required to obtain p-type semiconductor of 1 metric ton silicon. Concept introduction: The p-type semiconductor is obtained by doping an impurity having 3 valence electron like boron (B), Aluminum (Al) and gallium (Ga) to silicon (4 electron system). Thus, the covalent bond between the substrate and dopant always have one less electron, which causes a positive hole in the system and conductivity generated. However on the amount of dopant the silicon semiconductor may be subdivided into two category, light and heavy semiconductor. In light silicon semiconductor 1 impurity atom is present per 1,000,000,000 or ppb (parts per billion) silicon atoms. On the other hand for generation of heavy silicon semiconductor 1 atom of impurity needed per 1,000 atom of silicon.

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Answer 1

Question

Chapter 8, Problem 8.104PAE

Interpretation Introduction

(a)

Interpretation:

The mass of the dopant required to obtain p-type semiconductor of 1 metric ton silicon.

Concept introduction:

The p-type semiconductor is obtained by doping an impurity having 3 valence electron like boron (B), Aluminum (Al) and gallium (Ga) to silicon (4 electron system). Thus, the covalent bond between the substrate and dopant always have one less electron, which causes a positive hole in the system and conductivity generated. However on the amount of dopant the silicon semiconductor may be subdivided into two category, light and heavy semiconductor.

In light silicon semiconductor 1 impurity atom is present per 1,000,000,000 or ppb (parts per billion) silicon atoms. On the other hand for generation of heavy silicon semiconductor 1 atom of impurity needed per 1,000 atom of silicon.

Expert Solution

Answer to Problem 8.104PAE

Solution:

The amount of boron, the impurity,is required to generate light and heavy p-type 1 metric ton silicon semiconductor is 3.860×10−4 g and 386.093g respectively.

Explanation of Solution

1 metric ton of silicon = 106 gm of silicon.

On conversion of 106 gm of silicon to moles of silicon, 10628=3.571×104 moles (as molecular weight of silicon is 28).

1 mole of silicon is equivalent to 6.023×1023 number of atoms, thus 3.571×104 moles is equivalent to (6.023×1023×3.571×104)=2.151×1028 atoms of silicon.

For making light semiconductor 1 atom of dopant is required per 109 atoms of silicon.

Henceforth, number of dopant atoms required is 2.151×1028109=2.151×1019.

The numbers of moles of boron required as dopant are 2.151×10196.023×1023=3.571×10−5.

The mass of boron dopant required (3.571×10−5×10.811)=3.860×10−4 gm (As atomic weight of boron is 10.811g).

Henceforth to make the boron incorporated light p-type semiconductor per metric ton of silicon required 3.860×10−4 gm of boron.

To prepare heavy silicon semiconductor number of dopant atoms required is 2.151×1028103=2.151×1025.

The numbers of moles of boron required as dopant are 2.151×10256.023×1023=35.713.

The mass of boron dopant required (35.713×10.811)=386.093 gm.

Henceforth to make the boron incorporated heavy p-type semiconductor per metric ton of silicon required 386.093 gm of boron.

Interpretation Introduction

(b)

Interpretation:

The mole fraction of the dopant to obtain p-type silicon semiconductor is to be determined.

Concept introduction:

The mole number of the impurity or dopant present per unit of total mole number of the dopant and substrate in a semiconductor is called the mole fraction of the dopant. It can be expressed as-

The mole fraction of the dopant in semiconductor = The mole number of dopant presentThe mole number of dopant + mole number of substance

Expert Solution

Answer to Problem 8.104PAE

Solution:

The mole fraction of the dopant i.e. boron in p-type light and heavy silicon semiconductor is and respectively.

Explanation of Solution

For light silicon p-type semiconductor doped by boron the mole number of dopant and substrate are 3.571×10−5 and 3.571×104 respectively.

On plugging the values in the equation,

The mole fraction of the dopant in semiconductor = 3.571×10−5(3.571×10−5) + (3.571×104)

So, The mole fraction of the dopant = 3.571×10−53.571×104=1×10−9

Thus in the light semiconductor the mole fraction of the dopant is 1×10−9.

On the other side for heavy semiconductor doped by boron the mole number of dopant and substrate are 35.713 and 3.571×104 respectively.

On plugging the values in the equation,

The mole fraction of the dopant in semiconductor = 35.71335.713 + (3.571×104)

So, The mole fraction of the dopant = 35.7133.574×104=9.992×10−4

Thus in the light semiconductor the mole fraction of the dopant is 9.992×10−4.

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